Lecture 25 March 04, 2011
Ionic bonding and crystals
Nature of the Chemical Bond
with applications to catalysis, materials
science, nanotechnology, surface science,
bioinorganic chemistry, and energy
William A. Goddard, III, wag@wag.caltech.edu
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics,
California Institute of Technology
Teaching Assistants: Wei-Guang Liu <wgliu@wag.caltech.edu>
Caitlin Scott <cescott@caltech.edu>
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
1
Last time
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
2
Experimental data M2 phase – partial occupation
at M3, M4, M5, and Te1, Te2 sites
Formula: Mo4.31V1.36Te1.81Nb0.33O19.81
Unit cell Occupation
M3
M5
M5
M3
O
Te2
Te1
All V are VIV, there is no VV
DeSanto,.; Buttrey.; Grasselli,.; Lugmair.; Volpe.; Toby.; Vogt, Z. Kristallographie 2004, 219, 152-165.
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
3
Initial Structure M2 phase
To resolve partial occupations
[010]
•
•
•
•
2x3x4 super cell
25.26Å x 21.88Å x 16.08Å
28*24 = 672 atoms
10 different initial structures
•
•
•
•
Te : 48
M3: Mo=13/V=11
M4: Mo=75/V=21
M5: Mo=24
[100]
Num/unitcel
Occupation
Element
l
Ratio
M1/M2
~2
Te1/Te2 0.237/0.218
Positions
M3
M4
M5
Mo
V(Nb)
23
21
1
Mo/V 0.54/0.46 13
11
4
Mo/V 0.78/0.22 75
21
1 27 configurations,
Mo/Nb
- Use24
1.95x10
Monte Carlo
Ch120a-Goddard-L25
Total
error
44
24
96
24
0.74%
0.67%
0.73%
1.54%4
© copyright 2011 William A. Goddard III, all rights reserved
Monte Carlo Swap
Simulation 400,000 MC-steps
Displacement temperature: 500K
Displacement Details:
•
•
•
•
step-size: 0.1 Å
MC-swap temperature: 25000K
Swap energy bias: 10.0 kcal/mol
Swap frequency: 5
Analysis Method
(001) face
[001]
[010]
Each Vertical Column
[100]
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
5
Conformation of Te-O chain
Te1 BVS:3.372
Te2 BVS:4.119
No Vanadium
M5
M3
M5
M5
M3
Ch120a-Goddard-L25
Te1
Te2
M3
M3
M5
Tellurium has zigzag
conformation in the center of
channel
© copyright 2011 William A. Goddard III, all rights reserved
6
Distribution of Nb in M5 sites
have 6 M5 sites in 2x3 supercell
8 Nb (open) and 16 Mo (filled)
z
M3
M5
M5
1
Nb column
M3
2
3
4
5
Mo
Ch120a-Goddard-L25
Nb prefer to stay one column
Nb
© copyright
2011 William A. Goddard III, all rights reserved
7
Distribution of V,Mo in M3 and M4 sites
when M5 site is Nb (not Mo)
Types of Vertical Distribution
A Red  VVVV
B Orange  VVVMo
y
x
C Green  VVMoMo
E Blue  VMoMoMo
F Purple  MoMoMoMo
Mo
Mo
Nb
Mo
Mo
Mo
Mo prefers nucleate VVVV
Nb decreases the fraction of VVVV
May be best for activation
May decrease activity for propane
8
of propane
© copyright 2011 William A. Goddard
III, all rights reserved
ButCh120a-Goddard-L25
may increase selectivity
QM calculation on Te-O chain, replacing in-plane Te-O-M bonds
with Te-OH
OH
O
HO
Te
HO
HO
Te
HO
O
O
OH
OH
Te
Te
OH
OH
O
HO
O
HO
Te
HO
Te
O
HO
OH
Te
O
Te
OH
OH
HO
OH
H2O
TeF4 is hypervalent with in-plane bonds at ~95° and out-of-plane
bonds at ~180°.
copyright 2011 William A. Goddard
III, all rights reserved
ForCh120a-Goddard-L25
M2 Phase we find© Te-O-Te-O-Te-O
hypervalent
chains
9
4
-31.1 -25
(-27.2)
[-1.4]
-40
-17
(-15
[8
-19.9
2
Expect that Te=O
can
extract
allylic
Hydrogen
(-16.3)
-39.7
-55
C
-70
HN
H
O
Mo
HO
Te
HO
-85
O
H
H OO
Mo O
O
O
O
OH
Te
OH
O
HO
(-35.0)
[-9.4]
Te
HO
O
Mo
O
All Te are
TeIV before
and after
the H
abstraction
C H
O
H N
OH
O Mo O
Mo
Mo
OH
O
O
O
HO
Te
HO
O
Te
HO
C HO
O
Mo O OH2
-55
O
H
Mo
Mo
C
O
H OOO O
O
HN
Mo O
O
O
-70
Mo
Mo
OH
O
O
O
OH
N
Te
HO
O
Te
OH
HO
OH
Te=O extracts
Allylic H to from
TeOH while allyl
trapped on
nearby Mo=NH
H2O
Ch120a-Goddard-L25
H
C
-39.7
(-35.0)
N
O
[-9.4]
Mo
-85
O
O
OH
2
O
OH
Te
[16.0]
-40
Continue with previous mechanism
for propene on BiMoOx
10
© copyright 2011 William A. Goddard III, all rights reserved
Surface of M2
Build Process
Start from the final M2 crystal obtained from
ReaxFF-MC (2x3x4)
Cleave a surface on (001) face/Saturated
with oxygen/Build a vacuum slab/MM
Top and Bottom surface oxygen
The final surface contains 71
oxygen atoms(74.0% coverage),
32 on the top and 39 on the
bottom (totally 96 oxygen
positions)
Ch120a-Goddard-L25
Remove oxygens on the top or bottom
surface lower than 1/2O2 energy
Do the cycle of MM and Remove
Get the Surface
© copyright 2011 William A. Goddard III, all rights reserved
11
Active site for propene activation
Use H atom to test activity of each site of the surface
H added to the surface
TeO is the most active
site of M2 phase for
abstracting Hydrogen
from propene
Ch120a-Goddard-L25
H-allyl bond energy:-88.0kcal/mol
TeO—H
© copyright 2011 William A. Goddard III, all rights reserved
12
Active center
O
O
O
Mo O
O
O
Te
O
Mo
O
O
Mo
O
V
O
O
V
O
O
Mo
O
O
V
O
O
+
+
O
Te
+
O
O
O
Mo
O
Te
O
O
Mo
O
O
vertical oxygen are hidden
M4
M5
M4
M4
M3
M4
M4
M4
M5
Ch120a-Goddard-L25
• Active site can contain one V, two V or three
V
• one VIV  one TeIV exists in the channel
2V  2Te, 3V  3Te
• M5 definitely would contain some NbV if
there2011
areWilliam
onlyA. one
orIII,two
VIVreserved
in the center 13
© copyright
Goddard
all rights
Conclusion about M2 phase
• Vanadium prefers to stay as one column in M3
and M4 sites
• Nb has effect of segregating the Vanadium
columns
• Te-O in the center of channel in zigzag
conformation, TeIV
• Surface coverage of oxygen on M2 surface is
74.0%
• TeO hypervalent chain is the most active site
for abstracting allyic H from propene
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
14
Reactions of hydrocarbons on Ni468 nanoparticle
Jan. 20, 2010
New paper on ReaxFF
6 cases:
120 methane,
60 ethene,
60 ethyne,
40 propene,
20 benzene,
20 Cylclohexane
Initial and Final structures for ReaxFF RD simulation of 40 propene molecules
adsorbing and decomposing on a Ni468 cluster
Ch120a-Goddard-L25
Ni ©468
particle, 21A diameter
copyright 2011 William A. Goddard III, all rights reserved
15
ReaxFF: Acetylene Adsorption & Decomposition on Ni468
nanoparticle
Start: 60 C2H2
end:
52 Cad + 2 C2H3 gas + 2 C2H2ad + C2Had+C2ad
Ch120a-Goddard-L25
Conclusions
1. Both C-H bonds
break before the C-C
bond breaks
2. Formation of
subsurface C helps
break C-C bonds. 17
© copyright 2011 William A. Goddard III, all rights reserved
Ethyne detail
Reaction of C with 2nd
layer Ni very important
Build up surface NixCx in
first few rows
Dynamics of surface
Ni plays important
role in dissociating C2
Get some carbon into
interior
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
18
ReaxFF: Benzene Adsorption
& Decomposition on Ni
Particle
C 6 Hx
H2
C2
C6H6ad
Simplified sequence
C6H6C6H5C6H4C6H3C5H3
C5H2C4H2C4HC3HC3
C2C
At the end
7 Ch120a-Goddard-L256
46 2011 William A. Goddard III, all rights reserved
© copyright
19
Benzene detail
Benzene chemisorbs
horizontally on the Ni
C6H6 chemisorbed
C6H3-allyl
particle surface
chemisorbed
through pi electrons.
As H removed, get
strong C-Ni sigma
bonds, reorienting
benzene vertically.
C atoms denuded of H
are “swallowed” by
the particle by PacC3H with bare C in
C6H3-allyl tail in surface
Man mechanism, for
subsurface
cleaving C-C bonds.
C-H bonds far from the
surface are
protected until the C
atoms separating
them from the
surface are “eaten”
away.
20
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
Early Stages of CNT Growth from Acetylene Feedstock at
1500K on Ni468 nanoparticle (21A)
2 nanosec
NVT-RD
Start with 100 gas phase
C2H2 molecules, add an
additional 50 molecules
every 200 ps. At end 350
C 2H 2
Ch120a-Goddard-L25
• NVT-RD
• 1 nanosecond +
• 1000 K, 1500 K,
2000K, 2500 K 0.5 fs
time- step
• 100 fs T-damping
© copyright 2011 William A. Goddard III, all rights reserved
21
CNT Nucleation Study (after 2ns ReaxFF RD)
At the end of the
simulation we are left
with a large carbon
ring structure (C367H78)
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
22
Subsurface Analysis of Acetylene Feedstock
Decomposition on Ni nanoparticle
300 acetylene molecules after 2 ns RD at 2500K
radial atom distribution after 2ns RD at 2500K
Cross section
side view
140
# of atoms
120
Ni
100
C
80
H
60
40
20
15
.5
14
12
.5
11
9.
5
8
6.
5
5
3.
5
2
0.
5
0
radial distance (Angstroms)
C atoms penetrate 10.5A to the core of the catalyst
particle forming nickel carbide
Cross section head on
H penetrates
only part way© copyright
in, preferring
theA.surface.
Ch120a-Goddard-L25
2011 William
Goddard III, all rights reserved
23
Experimental Confirmation of a Yarmulke
Mechanism
Atomic-scale, video-rate environmental transmission
microscopy was used to monitor the nucleation and
growth of single walled nanotubes.
Ch120a-Goddard-L25
Hofmann,
S.2011
et al.
Nano
2007,
602.
© copyright
William
A. Lett.
Goddard
III, all7,
rights
reserved
24
New material
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
25
Ionic bonding (chapter 9)
Consider the covalent bond of Na to Cl. There Is very little
contragradience, leading to an extremely weak bond.
Alternatively, consider
transferring the charge
from Na to Cl to form
Na+ and ClCh120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
26
The ionic limit
At R=∞ the cost of forming Na+ and Clis IP(Na) = 5.139 eV minus EA(Cl) = 3.615 eV = 1.524 eV
But as R is decreased the electrostatic energy drops as
DE(eV) = - 14.4/R(A) or DE (kcal/mol) = -332.06/R(A)
Thus this ionic curve crosses the covalent curve at
R=14.4/1.524=9.45 A
Using the bond distance
of NaCl=2.42A leads to
a coulomb energy of
6.1eV leading to a bond
of 6.1-1.5=4.6 eV
The exper De = 4.23 eV
Showing that ionic
character dominates
Ch120a-Goddard-L25
E(eV)
© copyright 2011 William A. Goddard III, all rights reserved
R(A)
27
GVB orbitals
of NaCl
Dipole moment
= 9.001 Debye
Pure ionic
11.34 Debye
Thus
Dq=0.79 e
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
28
electronegativity
To provide a measure to estimate polarity in bonds, Linus
Pauling developed a scale of electronegativity () where
the atom that gains charge is more electronegative and
the one that loses is more electropositive
He arbitrarily assigned
=4 for F, 3.5 for O, 3.0 for N, 2.5 for C, 2.0 for B, 1.5 for
Be, and 1.0 for Li
and then used various experiments to estimate other
cases . Current values are on the next slide
Mulliken formulated an alternative scale such that
M= (IP+EA)/5.2
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
29
Electronegativity
Based on M++
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
30
Comparison of Mulliken and Pauling electronegativities
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
31
Ionic crystals
Starting with two NaCl monomer, it is downhill by 2.10
eV (at 0K) for form the dimer
Because of repulsion between
like charges the bond lengths,
increase by 0.26A.
A purely electrostatic calculation would
have led to a bond energy of 1.68 eV
Similarly, two dimers can combine to form
a strongly bonded tetramer with a nearly
cubic structure
Continuing, combining 4x1018 such
dimers leads to a grain of salt in which
each Na has 6 Cl neighbors and each Cl
hasCh120a-Goddard-L25
6 Na neighbors © copyright 2011 William A. Goddard III, all rights reserved
32
The NaCl or B1 crystal
All alkali halides
have this
structure except
CsCl, CsBr, CsI
(they have the B2
structure)
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
33
The CsCl or B2 crystal
There is not yet a good understanding of the fundamental
reasons why particular compound prefer particular
structures. But for ionic crystals the consideration of ionic
radii has proved useful
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
34
Ionic radii, main group
Fitted to various crystals. Assumes O2- is 1.40A
NaCl R=1.02+1.81 = 2.84, exper is 2.84
From
R. D. Shannon, Acta©Cryst.
751 (1976)
Ch120a-Goddard-L25
copyrightA32,
2011 William
A. Goddard III, all rights reserved
35
Ionic radii, transition metals
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
36
Ionic radii Lanthanides and Actinide
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
37
Role of ionic sizes in determining crystal structures
Assume that the anions are large and packed so that they
contact, so that 2RA < L, where L is the distance between anions
Assume that the anion and cation are in contact.
Calculate the smallest cation consistent with 2RA < L.
RA+RC = L/√2 > √2 RA
RA+RC = (√3)L/2 > (√3) RA
Thus RC/RA > 0.414
Thus RC/RA > 0.732
Thus for 0.414 < (RC/RA ) < 0.732 we expect B1
For (RC/RA ) > 0.732 either is ok.
ForCh120a-Goddard-L25
(RC/RA ) < 0.414 must
be2011
some
other
structure
© copyright
William A.
Goddard
III, all rights reserved
38
Radius Ratios of Alkali Halides and Noble metal halices
Rules work ok
B1: 0.35 to 1.26
B2: 0.76 to 0.92
Based on R. W.
G. Wyckoff,
Crystal
Structures, 2nd
edition. Volume 1
(1963)
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
39
Wurtzite or B4 structure
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
40
Sphalerite or Zincblende or B3 structure GaAs
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
41
Radius rations B3, B4
The height of the tetrahedron is (2/3)√3 a where a is the side of
the circumscribed cube
The midpoint of the tetrahedron (also the midpoint of the cube) is
(1/2)√3 a from the vertex.
Hence (RC + RA)/L = (½) √3 a / √2 a = √(3/8) = 0.612
Thus 2RA < L = √(8/3) (RC + RA) = 1.633 (RC + RA)
Thus 1.225 RA < (RC + RA) or RC/RA > 0.225
Thus B3,B4 should be the stable structures for
0.225 < (RC/RA) < 0. 414
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
42
Structures for II-VI compounds
B3 for 0.20 < (RC/RA) < 0.55
B1 for 0.36 < (RC/RA) < 0.96
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
43
CaF2 or fluorite structure
Like GaAs but
now have F at
all tetrahedral
sites
Or like CsCl
but with half
the Cs missing
Find for RC/RA > 0.71
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
44
Rutile (TiO2) or Cassiterite (SnO2) structure
Related to NaCl
with half the
cations missing
Find for RC/RA < 0.67
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
45
CaF2
rutile
CaF2
rutile
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
46
Electrostatic Balance Postulate
For an ionic crystal the charges transferred from all cations
must add up to the extra charges on all the anions.
We can do this bond by bond, but in many systems the
environments of the anions are all the same as are the
environments of the cations. In this case the bond polarity
(S) of each cation-anion pair is the same and we write
S = zC/nC where zC is the net charge on the cation and nC is
the coordination number
Then zA = Si SI = Si zCi /ni
Example1 : SiO2. in most phases each Si is in a tetrahedron
of O2- leading to S=4/4=1.
Thus each O2- must have just two Si neighbors
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
47
a-quartz structure of SiO2
Each Si bonds to 4 O,
OSiO = 109.5°
each O bonds to 2 Si
Si-O-Si = 155.x °
Helical chains
single crystals optically active;
α-quartz converts to β-quartz
at 573 °C
From wikipedia
Ch120a-Goddard-L25
rhombohedral
(trigonal)
hP9, P3121
No.152[10]
© copyright 2011 William A. Goddard III, all rights reserved
48
Example 2 of electrostatic balance: stishovite phase of SiO2
The stishovite phase of SiO2 has six coordinate Si,  S=2/3.
Thus each O must have 3 Si neighbors
Rutile-like structure, with 6coordinate Si;
high pressure form
densest of the SiO2
polymorphs
From wikipedia
Ch120a-Goddard-L25
tetragonal
tP6, P42/mnm,
No.136[17]
© copyright 2011 William A. Goddard III, all rights reserved
49
TiO2, example 3 electrostatic balance
Example 3: the rutile, anatase, and brookite phases of TiO2
all have octahedral Ti.
Thus S= 2/3 and each O must be coordinated to 3 Ti.
top
anatase phase TiO2
front
Ch120a-Goddard-L25
right
© copyright 2011 William A. Goddard III, all rights reserved
50
Corundum (a-Al2O3). Example 4 electrostatic balance
Each Al3+ is in a distorted octahedron,
leading to S=1/2.
Thus each O2- must be coordinated to 4 Al
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
51
Olivine. Mg2SiO4. example 5 electrostatic balance
Each Si has four O2- (S=1) and each
Mg has six O2- (S=1/3).
Thus each O2- must be coordinated to
1 Si and 3 Mg neighbors
O = Blue atoms (closest packed)
Si = magenta (4 coord) cap voids in
zigzag chains of Mg
Mg = yellow (6 coord)
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
52
Illustration, BaTiO3
A number of important oxides have the perovskite structure
(CaTiO3) including BaTiO3, KNbO3, PbTiO3.
Lets try to predict the structure without looking it up
Based on the TiO2 structures , we expect the Ti to be in an
octahedron of O2-, STiO = 2/3.
How many Ti neighbors will each O have?
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
53
Illustration, BaTiO3
A number of important oxides have the perovskite structure
(CaTiO3) including BaTiO3, KNbO3, PbTiO3.
Lets try to predict the structure without looking it up
Based on the TiO2 structures , we expect the Ti to be in an
octahedron of O2-, STiO = 2/3.
How many Ti neighbors will each O have?
It cannot be 3 since there would be no place for the Ba.
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
54
Illustration, BaTiO3
A number of important oxides have the perovskite structure
(CaTiO3) including BaTiO3, KNbO3, PbTiO3.
Lets try to predict the structure without looking it up
Based on the TiO2 structures , we expect the Ti to be in an
octahedron of O2-, STiO = 2/3.
How many Ti neighbors will each O have?
It cannot be 3 since there would be no place for the Ba.
It is likely not one since Ti does not make oxo bonds.
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
55
Illustration, BaTiO3
A number of important oxides have the perovskite structure
(CaTiO3) including BaTiO3, KNbO3, PbTiO3.
Lets try to predict the structure without looking it up
Based on the TiO2 structures , we expect the Ti to be in an
octahedron of O2-, STiO = 2/3.
How many Ti neighbors will each O have?
It cannot be 3 since there would be no place for the Ba.
It is likely not one since Ti does not make oxo bonds.
Thus we expect each O to have two Ti neighbors, probably at
180º. This accounts for 2*(2/3)= 4/3 charge.
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
56
Illustration, BaTiO3
A number of important oxides have the perovskite structure
(CaTiO3) including BaTiO3, KNbO3, PbTiO3.
Lets try to predict the structure without looking it up
Based on the TiO2 structures , we expect the Ti to be in an
octahedron of O2-, STiO = 2/3.
How many Ti neighbors will each O have?
It cannot be 3 since there would be no place for the Ba.
It is likely not one since Ti does not make oxo bonds.
Thus we expect each O to have two Ti neighbors, probably at
180º. This accounts for 2*(2/3)= 4/3 charge.
Now we must consider how many O are around each Ba, nBa,
leading to SBa = 2/nBa, and how many Ba around each O,
nOBa.
Ch120a-Goddard-L25
© copyright 2011 William A. Goddard III, all rights reserved
57
Illustration, BaTiO3
A number of important oxides have the perovskite structure
(CaTiO3) including BaTiO3, KNbO3, PbTiO3.
Lets try to predict the structure without looking it up
Based on the TiO2 structures , we expect the Ti to be in an
octahedron of O2-, STiO = 2/3.
How many Ti neighbors will each O have?
It cannot be 3 since there would be no place for the Ba.
It is likely not one since Ti does not make oxo bonds.
Thus we expect each O to have two Ti neighbors, probably at
180º. This accounts for 2*(2/3)= 4/3 charge.
Now we must consider how many O are around each Ba, nBa,
leading to SBa = 2/nBa, and how many Ba around each O,
nOBa.
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Prediction of BaTiO3 structure : Ba coordination
Since nOBa* SBa = 2/3, the missing charge for the O, we have
only a few possibilities:
nBa= 3 leading to SBa = 2/nBa=2/3 leading to nOBa = 1
nBa= 6 leading to SBa = 2/nBa=1/3 leading to nOBa = 2
nBa= 9 leading to SBa = 2/nBa=2/9 leading to nOBa = 3
nBa= 12 leading to SBa = 2/nBa=1/6 leading to nOBa = 4
Each of these might lead to a possible structure.
The last case is the correct one for BaTiO3 as shown.
Each O has a Ti in the +z and –z directions plus four Ba
forming a square in the xy plane
The Each of these Ba sees 4 O in the xy plane, 4 in the xz
plane and 4 in the yz plane.
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59
BaTiO3 structure (Perovskite)
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60
How estimate charges?
We saw that even for a material as ionic as NaCl diatomic, the
dipole moment  a net charge of +0.8 e on the Na and -0.8 e
on the Cl.
We need a method to estimate such charges in order to
calculate properties of materials.
First a bit more about units.
In QM calculations the unit of charge is the magnitude of the
charge on an electron and the unit of length is the bohr (a0)
Thus QM calculations of dipole moment are in units of ea0 which
we refer to as au. However the international standard for
quoting dipole moment is the Debye = 10-10 esu A
Where m(D) = 2.5418 m(au)
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61
Fractional ionic character of diatomic molecules
Obtained from the experimental dipole moment in Debye, m(D), and bond
distance R(A) by dq = m(au)/R(a0) = C m(D)/R(A) where C=0.743470. Postive 
62
© copyright 2011 William A. Goddard III, all rights reserved
thatCh120a-Goddard-L25
head of column is negative
Charge Equilibration
First consider how the energy of an atom depends on
the net charge on the atom, E(Q)
Including terms through 2nd order leads to
Charge Equilibration for Molecular
Dynamics Simulations;
A. K. Rappé and W. A. Goddard III;
J. Phys. Chem. 95, 3358 (1991)
(2)
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(3)
63
Charge dependence of the energy (eV) of an atom
E=12.967
Harmonic fit
E=0
E=-3.615
Cl+
Q=+1
Cl
Q=0
Ch120a-Goddard-L25
Cl-
Q=-1
= 8.291
Get minimum at Q=-0.887
Emin = -3.676
= 9.352
© copyright 2011 William A. Goddard III, all rights reserved
64
QEq parameters
Ch120a-Goddard-L25
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65
Interpretation of J, the hardness
Define an atomic radius as
RA0
Re(A2) Bond distance of
homonuclear
H
0.84 0.74
diatomic
C
1.42 1.23
N
1.22 1.10
O
1.08 1.21
Si
2.20 2.35
S
1.60 1.63
Li
3.01 3.08
Thus J is related to the coulomb energy of a charge the size of the
66
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atom
The total energy of a molecular complex
Consider now a distribution of charges over the
atoms of a complex: QA, QB, etc
Letting JAB(R) = the Coulomb potential of unit
charges on the atoms, we can write
Taking the derivative with respect to charge leads to the
chemical potential, which is a function of the charges
or
The definition of equilibrium is for all chemical potentials to be
equal.
This leads to © copyright 2011 William A. Goddard III, all rights reserved
Ch120a-Goddard-L25
67
The QEq equations
Adding to the N-1 conditions
The condition that the total charged is fixed (say at 0)
leads to the condition
Leads to a set of N linear equations for the N variables QA.
AQ=X, where the NxN matrix A and the N dimensional vector A
are known. This is solved for the N unknowns, Q.
We place some conditions on this. The harmonic fit of charge to
the energy of an atom is assumed to be valid only for filling the
valence shell.
Thus we restrict Q(Cl) to lie between +7 and -1 and
Q(C) to be between +4 and -4
Similarly Q(H) is between +1 and -1
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68
The QEq Coulomb potential law
We need now to choose a form for JAB(R)
A plausible form is JAB(R) = 14.4/R, which is valid when the
charge distributions for atom A and B do not overlap
Clearly this form as the problem that JAB(R)  ∞ as R 0
In fact the overlap of the orbitals leads to shielding
The plot shows the
shielding for C atoms using
various Slater orbitals
And l = 0.5
Ch120a-Goddard-L25
Using RC=0.759a0
© copyright 2011 William A. Goddard III, all rights reserved
69
QEq results for alkali halides
Ch120a-Goddard-L25
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70
QEq for Ala-His-Ala
Amber
charges in
parentheses
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71
QEq for deoxy adenosine
Amber
charges in
parentheses
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72
QEq for polymers
Nylon 66
PEEK
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73
Perovskites
Perovskite (CaTiO3) first described in the 1830s
by the geologist Gustav Rose, who named it
after the famous Russian mineralogist Count Lev
Aleksevich von Perovski
crystal lattice appears cubic, but it is actually
orthorhombic in symmetry due to a slight
distortion of the structure.
Characteristic chemical formula of a perovskite
ceramic: ABO3,
A atom has +2 charge. 12 coordinate at the
corners of a cube.
B atom has +4 charge.
Octahedron of O ions on the faces of that cube
centered on a B ions at the center of the cube.
Together A and B form an FCC structure
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74
The stability of the perovskite structure depends on
the relative ionic radii:
Ferroelectrics
if the cations are too small for close packing with the
oxygens, they may displace slightly.
Since these ions carry electrical charges, such
displacements can result in a net electric dipole
moment (opposite charges separated by a small
distance).
The material is said to be a ferroelectric by analogy
with a ferromagnet which contains magnetic dipoles.
At high temperature, the small green B-cations can
"rattle around" in the larger holes between oxygen,
maintaining cubic symmetry.
A static displacement occurs when the structure is
cooled below the transition temperature.
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75
Phases of BaTiO3
<111> polarized
rhombohedral
<110> polarized
orthorhombic
-90oC
<100> polarized
tetragonal
120oC
5oC
Non-polar
cubic
Temperature
Different phases of BaTiO3
Ba2+/Pb2+
c
Ti4+
O2-
a
Non-polar cubic
above Tc
Six variants at room temperature
c
 1.01 ~ 1.06
a
<100> tetragonal
below Tc
Domains separated by domain walls
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76
Nature of the phase transitions
Displacive model
Assume that the atoms prefer to
distort toward a face or edge or
vertex of the octahedron
Increasing
Temperature
Different phases of BaTiO3
<111> polarized
rhombohedral
<110> polarized
orthorhombic
-90oC
face
<100> polarized
tetragonal
120oC
5oC
edge
Non-polar
cubic
vertex
Temperature
center
1960 Cochran
Soft Mode Theory(Displacive Model)
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77
Nature of the phase transitions
Displacive model
Assume that the atoms prefer to
distort toward a face or edge or
vertex of the octahedron
1960
Cochran
Increasing
Temperature
Soft Mode Theory(Displacive Model)
Order-disorder
1966
Bersuker
Eight Site Model
1968
Comes
Order-Disorder Model (Diffuse X-ray Scattering)
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78
Comparison to experiment
Displacive  small latent heat
This agrees with experiment
R  O: T= 183K, DS = 0.17±0.04 J/mol
O  T: T= 278K, DS = 0.32±0.06 J/mol
T  C: T= 393K, DS = 0.52±0.05 J/mol
Diffuse xray scattering
Expect some disorder,
agrees with experiment
Cubic
Tetra.
Ortho.
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Rhomb.
79
Problem displacive model: EXAFS & Raman observations
d
(001)
EXAFS of Tetragonal Phase[1]
•Ti distorted from the center of oxygen octahedral in tetragonal
phase.
α
(111)
•The angle between the displacement vector and (111) is α= 11.7°.
Raman Spectroscopy of Cubic Phase[2]
A strong Raman spectrum in cubic phase is found in experiments.
But displacive model  atoms at center of octahedron: no Raman
1.
B. Ravel et al, Ferroelectrics, 206, 407 (1998)
2.
A. M. Quittet et al, Solid
State Comm.,
12, 1053
(1973) III, all rights reserved
Ch120a-Goddard-L25
© copyright
2011 William
A. Goddard
80
80
QM calculations
The ferroelectric and cubic phases in BaTiO3 ferroelectrics are also
antiferroelectric
Zhang QS, Cagin T, Goddard WA
Proc. Nat. Acad. Sci. USA, 103 (40): 14695-14700 (2006)
Even for the cubic phase, it is lower energy for the Ti
to distort toward the face of each octahedron.
How do we get cubic symmetry?
Combine 8 cells together into a 2x2x2 new unit cell,
each has displacement toward one of the 8 faces, but
they alternate in the x, y, and z directions to get an
overall cubic symmetry
Microscopic Polarization
Ti atom
distortions
Cubic
I-43m
Ch120a-Goddard-L25 z
=
Pz
Py
Px
+
Macroscopic
Polarization
+
© copyright 2011 William A. Goddard III, all rights reserved
=
81
QM results explain EXAFS & Raman observations
d
(001)
EXAFS of Tetragonal Phase[1]
•Ti distorted from the center of oxygen octahedral in tetragonal
phase.
α
(111)
•The angle between the displacement vector and (111) is α= 11.7°.
PQEq with FE/AFE model gives α=5.63°
Raman Spectroscopy of Cubic Phase[2]
A strong Raman spectrum in cubic phase is found in experiments.
1.
Model
Inversion symmetry in
Cubic Phase
Raman Active
Displacive
Yes
No
FE/AFE
No
Yes
B. Ravel et al, Ferroelectrics, 206, 407 (1998)
2.
A. M. Quittet et al, Solid
State Comm.,
12, 1053
(1973) III, all rights reserved
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2011 William
A. Goddard
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